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Partial derivate philosophy

  1. Aug 17, 2006 #1
    partial derivate "philosophy"

    I'm having problems with fundementals of partial derivatives. I've looked around for something like "partial derivative FAQ", but couldn't find.

    Can any of the following be true in general?

    [tex]\frac{\partial f}{\partial x} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}[/tex]
    [tex]\frac{d f}{d x} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}[/tex]

    I'd rather expect
    [tex]\frac{d f}{d x} = \frac{d f}{d g}\frac{d g}{d x}[/tex]
    but the partial version seems to be working equally. Can there be cases where the partial version fails?

    Also, it's weird that although
    [tex]\frac{d f}{d g} = \frac{1}{\frac{d g}{d f}}[/tex]
    holds for all,
    [tex]\frac{\partial f}{\partial g} = \frac{1}{\frac{\partial g}{\partial f}}[/tex]
    doesn't. Why not?

    I'm desperately looking for a proof for any of them...

    p.s.: You may ask which one is independent variable and on what f and g depends. I leave it to the guy who's willing to prove them, because I'm not sure whether if makes any difference in a general proof.
     
    Last edited: Aug 17, 2006
  2. jcsd
  3. Aug 17, 2006 #2
    You might want to look up the chain rule(s) for partial derivatives because I think I remember covering a few of these cases in my Calc 3 class.
     
  4. Aug 21, 2006 #3

    radou

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    Homework Helper

    Step 1: analyse the function and see if it's a composition of functions; Step 2: remember that a total differential of a function and a partial derivation are not the same; Step 3: apply rules.
     
  5. Sep 4, 2006 #4
    suppose you have a function
    F(x,y)=x^3+y^3
    if we want partial derivative of Fx
    we write
    fx(x, y)= 3x^2
    if we want partial derivative of Fy
    we write
    fy(x,y)= 3y^2
     
  6. Sep 4, 2006 #5
    I think that the original poster already knew this, and it's not a matter of "writing" it's a matter opf taking the derivative with respect to a single variable in a function of several variables.
     
  7. Sep 5, 2006 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You can't prove anything until you know what it is you are trying to prove. And that depends on just the things you are not telling us!

    When you write
    [tex]\frac{\partial f}{\partial x} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}[/tex]
    the fact that you are using partial fractions implies that f is a function of several variables, one of which is g and we don't know what the others are, that g is itself a function of several variables, one of which is x. We could assume that the other variables f depends on do not themselves depend on x- so those partial derivatives are 0. If g and those other unknown variables depend on other variables, say y and z, then what you have makes sense: f can be written as a function of x, y, z, say and we are calculating the derivative of f with respect to x only.

    On the other hand
    [tex]\frac{d f}{d x} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}[/tex]
    I don't think can make sense. If f depends on g, whether there are other variables or not, and g depends on x as well as other variables, implied by the [itex]\frac{\partial g}{\partial x}[/itex] then f must depend on those other variables as well and df/dx makes no sense.
     
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